光纤通信技术课后部分习题答案Fiber-OpticCommunicationsTechnology

Sep.26thCalculatetheenergyofasinglephotonat1300nmandat1500nm.Solution:Asinglephotonat1300nm:E=hf=chi1=3·108.6.626·10_34=1.529-10-19(J)p1300-10-9Asinglephotonat1550nm:E=hf=chiA=3·108.6.626·10-34=1.282-10-1\J)P1550-10-92.7a.Howmanyphotonspersecondemanatefromalaserdioderadiatingat1300nmifitspoweris1mW?b.Solution:a:b:Ps13.12Solution:Howmanyphotonspersecondemanateat1550nm?Usingtheresultabove:E=Pt=1-10-3·1=10-3(J)E=E·NpE=E·NpN=..E.._=w-3=6.540·1015EP1.529-10-19N=..E.._=10-3=7.798·1015EP1.282·10-19Foraspecificfiber,NA=0.2375andn1=1.4860.Findn2(ncladding).hence0.23752=1.48602-niHencen2=1.46693.22Anopticalfiberwithattenuationof0.25dB/kmisusedfor20-krntransmission.Thelightpowerlaunchedintothefiberis2mW.Whatistheoutputpower?Solution:Loss(dB)=-10lg(Pout/Pin)A(dB/krn)=loss(dB)Ifiberlength(krn)-lQlgpor,tHence0.25=2mW'20Hence,Pout=0.6325mW.3.25Findthemaximumtransmissiondistanceforafiberlinkwithanattenuationof0.3dB/kmifthepowerlaunchedinis3mWandthereceiversensitivityis100µW.Solution:Loss(dB)=-10lg(Pout/Pin)A(dB/km)=loss(dB)Ifiberlength(km)-lOlg100.10-33Hence,0.3=------fiberlengthHence,fiberlength=49.24km3.7Thecorerefractiveindexis1.4513andthecladdingindexis1.4468.Whatis(1)thecriticalpropagationangle?(2)theacceptanceangle?(3)thenumericalaperture?Solution:(1)Thecriticalangleis:.-•ffi)'(n2)2.-1a=sm--=smn11-(1.4468)2=4.513101.4513(2):Theacceptanceangleis:Wa=2sin-1(n1sinaJ=2sin-1(1.4513·sin4.5131°)=13.1148°(3):TheNA:NA=sin0a=0.11423.21Whatdoesthetermtransparentwindowsmean?Specifythreepeakwavelengthsforthetransparentwindowsinmodemopticalfibers.Solution:Thetermtransparentwindowsmeansthattheregionswhereabsorptionreachesminimumvalueinthetypicalspectralattenuationdiagram(Loss(dB/km)-A(nm)diagram).Thethreepeakwavelengthsforthemare:850nm,1300nm,and1550nm.3.29Whatisthenumberofmodes(N.=V)foragraded-indexfiberifdis50µm,NAis0.200,andtheoperatingwavelengthis1300nm?Solution:TheVnumberis:v=trdNA=TC.50·10-6.0.200=24.2A,1300-10-9thenumberofmodesis:3.30Howmanymodescansupportastep-indexopticalfiberwhosed=8.3µm,n-Icore=1.4513,n2dad=1.4468,andA=1550nm?Solution:V=trdI2_2=.1r·8.3·10-6..J1.45132_1.44682=1.92A,v1lini1550-10-9V2.405,N=1P11E3.47Agraded-indexfiberhasn1=1.486andNA=0.200.Whatisthebitratefora1-kmlink?Solution:L·NA4L·NA4MG,=�---32cN332cn31I1BR-4MwhereNlcoregroupindexofrefractionHenceBR=8cn;=8.3·l08•l.4863=4.92·109bitIs=4.92GbitIs'GIL·NA41·103·0.2004Oct.17th4.30Agraded-indexfiberhasthefollowingcharacteristics:NA=0.200,dcore=50µmand').._,=1300nm.Whatpoweriscarriedbythefiber'scladding?S:V=7rClNA=z.50·l03·0.200=24.17,1,1300pclad=2J2=2J2=3_9%�otal3V3·24.17Hence,about3.9%ofthetotalpoweriscarriedbythefiber'scladding.4.51CalculatethepulsespreadingcausedbychromaticdispersionforBF04431-02multimodegrade-indexfiberfromSpecTran,(SeeFigure3.20)operatingat1300run.Assume!).')..,=70nm.S:TakingAo=1342.5nm,So=0.097ps/2km/nm·D(,1,)=SO(,1,_IL�)=0.())7.(Boo_1342.54J=-4_329ps/4;t3413003/nm·km!).tgmat=D(J)L'.t;t=4.329x70=303pslkm=0.3ns!km4.55Whatismeantbythetermdispersionpowerpenalty?CalculatethedispersonpowerpenaltyforthepulsespreadobtainedinProblem4.51.TheBRis2.5Gbit/s.S:Theamountoftransmittingpowerrequiredforcompensatingtheincreasebiterrorrate(BER)causebyattenuationiscalleddispersionpowerpenalty.InProblem4.51,/).1:to1a1=!).tmat=18ns(iffiberlengthis100km)Po(dB)=-10log10{exp[-(1/4)(!).tro1a1)2(1tBR)2]}=-10log10{exp[-1/4x(18x10-9)2(1tx2.5x109)2]}=2.17x104dB,enoughtoburnanytypeoffiber,impossible.ifthefiberlengthis1km:21t22P:=4.34-·t·BR4Po=2.17dB(realizable!)4.56Whatisthebit-ratelengthlimitationcausedbychromaticdispersonfortheSpectranfiberreferredtoinProblem4.51?S:(BRxL)max=1I[4crAD(11,)]=1/[4(LU/J2)D(?.)]=1/l4·70/J2·4.329xl0-12J=1.17Gb/s·kmChapter5-65.13Computethepulsespreadcausedbychromaticdispersionifafiberhasazero­dispersionwavelengthat1312nm,azero-dispersionslopeof0.090ps/2km,alengthof/nm·100km,andoperatesat1310nm.TheLD'sspectralwidthis1nm.S:D(2)=S0(2-20)=0.090·(1310-1312)=-0.18P:{,,m·kmMc1'=D(2)·&i·L=0.18·1·100=18ps5.17Calculatepulsespreadcausedbypolarization-modedispersionifDPMD=0.2psI.J,;;;i,andL=120km.S:l:!,.tPMD=DPMD•.Ji=0.2·m0=2.19ps6.2WhatistheV-numberoftheLow-PMDSMfromPlasmaOpticalFibreBV,assumingthatthecorediameterisequalto8.3µmandtherelativeindexequals0.36%?S:v=n:dn�=7r·8.3·10-6·1.467..Jo.36%=1.889k1215-10-96.8WhatlengthofDCFisneededtocompensateforchromaticdispersionata120-kmspanofLow-PMDfiberfromPlasmaOpticalFibreBV(seeFigure5.17)ifDocp(A)=-127ps/nm*kmandA=1550nm?S:6.10a.b.1:!,.tchrom=-1:!,.tcmnpensD(2)&tL4=-DDCF(2)&tLDCFs{i-�:}=-Doc,LDCF0.092(1550-l3l04)•120=127L.415503DCFL=16.50kmWhatmeansdowehavetocopewithPMD?Whatbitratecanbeachievedata120-kmspanwithLow-PMDFiberfromPlasmaOpticalFibreBV?Sa:WeusePMfiberorPMcomponentstogethertomakeaPMLink.WealsodevelopPMDcompensationtechniques.b:!itPMD=DPMD·.JL=0.2·MO=2.19ps11BRPMD===114GbitIs4/1tPMD4·2.19P2467.2AdryinggasflowsthroughapreformduringtheconsolidationphaseoftheOVDprocess.Whyisthisimportant?S:Itistheremoveresidualmoisturebecauseopticalfiber'sabsorptionpropertiesarecausedbysomemoleculesofthehydroxideani